Triangles (Similarity) — Class 10 Maths practice questions

Question 1

A school garden has a triangular flower bed ABC. The gardener places a wooden plank DE parallel to side BC, where D is a point on AB and E is a point on AC. The measurements are: AD = 3 m, DB = 6 m, and AE = 4 m.

(i) Find the length of EC. [1]
(ii) If BC = 9 m, find DE. [1]
(iii) The gardener wants to verify that △ADE ∼ △ABC. Prove that △ADE ∼ △ABC, clearly stating the similarity criterion used. [2]

Accepted Answer

(i) By the Basic Proportionality Theorem (BPT), if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, it divides the two sides in the same ratio.

Since DE ∥ BC,

AD/DB = AE/EC

⟹ 3/6 = 4/EC

⟹ EC = (4 × 6)/3

∴ EC = 8 m

(ii) Since DE ∥ BC, △ADE ∼ △ABC (AA similarity — established in part iii).

Corresponding sides of similar triangles are proportional:

DE/BC = AD/AB

AB = AD + DB = 3 + 6 = 9 m

⟹ DE/9 = 3/9

⟹ DE = 3

∴ DE = 3 m

(iii) Given: DE ∥ BC, D on AB, E on AC.

To Prove: △ADE ∼ △ABC.

Proof:

Since DE ∥ BC,

∠ADE = ∠ABC  (corresponding angles, DE ∥ BC with AB as transversal) …(i)

∠AED = ∠ACB  (corresponding angles, DE ∥ BC with AC as transversal) …(ii)

In △ADE and △ABC,

∠DAE = ∠BAC  (common angle) …(iii)

∠ADE = ∠ABC  [from (i)]

⟹ △ADE ∼ △ABC  (by AA similarity criterion)

Hence proved.

Question 2

A school garden has two triangular flower beds, △ABC and △PQR, on either side of a central path. The groundskeeper notes the following measurements:
• In △ABC: AB = 4 m, BC = 6 m, AC = 5 m
• In △PQR: PQ = 8 m, QR = 12 m, PR = 10 m
Based on this information, answer the following:
(i) What is the ratio AB : PQ? [1 mark]
(ii) Check whether △ABC ~ △PQR by verifying the ratio of all three pairs of corresponding sides. [1 mark]
(iii) If the area of △ABC is 9 m², find the area of △PQR. [2 marks]

Accepted Answer

(i) AB : PQ = 4 : 8 = 1 : 2
∴ AB : PQ = 1 : 2

(ii) Checking the ratio of all three pairs of corresponding sides:

AB/PQ = 4/8 = 1/2

BC/QR = 6/12 = 1/2

AC/PR = 5/10 = 1/2

⟹ AB/PQ = BC/QR = AC/PR = 1/2

Since all three pairs of corresponding sides are proportional, by the SSS Similarity Criterion,
∴ △ABC ~ △PQR

(iii) By the theorem on areas of similar triangles:
"The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides."

Area of △ABC / Area of △PQR = (AB/PQ)²

⟹ 9 / Area of △PQR = (1/2)²

⟹ 9 / Area of △PQR = 1/4

⟹ Area of △PQR = 9 × 4 = 36

∴ Area of △PQR = 36 m²

Question 3

A town planner is designing a triangular park. She places a lamp post at point D on side BC of triangular park ABC such that AD bisects angle BAC. The dimensions of the park are: AB = 6 m, AC = 9 m, BD = 4 m, and DC = 6 m.

(i) Verify whether AD is the angle bisector of ∠BAC using the Angle Bisector property. [1]
(ii) If a second triangular park PQR is to be built such that △ABC ∼ △PQR and PQ = 10 m, find the length of QR. [1]
(iii) The planner claims that △ABD ∼ △ACD. A colleague disagrees. Using the correct similarity criterion, determine whether △ABD ∼ △ACD. If yes, state the ratio of their areas. If no, justify your answer. [2]

Accepted Answer

(i) By the Angle Bisector Theorem, if AD bisects ∠BAC, then BD/DC = AB/AC.

LHS: BD/DC = 4/6 = 2/3

RHS: AB/AC = 6/9 = 2/3

∵ BD/DC = AB/AC = 2/3

∴ AD is the angle bisector of ∠BAC. (Verified) ✓

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(ii) Since △ABC ∼ △PQR, corresponding sides are proportional.

⟹ AB/PQ = BC/QR

BC = BD + DC = 4 + 6 = 10 m

⟹ 6/10 = 10/QR

⟹ QR = (10 × 10)/6 = 100/6

∴ QR = 50/3 m ≈ 16.67 m

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(iii) In △ABD and △ACD:

— AB = 6 m, AC = 9 m ⟹ AB ≠ AC
— BD = 4 m, DC = 6 m ⟹ BD ≠ DC
— AD is common to both triangles.

Checking SAS similarity: AB/AC = 6/9 = 2/3 and BD/DC = 4/6 = 2/3, but the included angle ∠ADB and ∠ADC are supplementary (∠ADB + ∠ADC = 180°), NOT equal. Therefore the SAS similarity criterion is NOT satisfied.

Checking AA similarity: ∠BAD = ∠CAD (since AD bisects ∠BAC). However, there is no second pair of equal angles that can be established without further information.

Checking SSS similarity: AB/AC = 2/3, BD/DC = 2/3, but AD/AD = 1 ≠ 2/3.

∵ No similarity criterion (AA, SAS, or SSS) is satisfied for △ABD and △ACD,

∴ △ABD is NOT similar to △ACD.

The planner's colleague is correct. The two triangles share the same height from A, so their areas are in the ratio of their bases:

Area(△ABD)/Area(△ACD) = BD/DC = 4/6 = 2/3

∴ The ratio of areas of △ABD to △ACD = 2 : 3.

(Note: The triangles are not similar, but their area ratio = 2 : 3 since they share the same vertex A and their bases BD and DC lie on the same line BC.)

Question 4

A city park has a triangular fountain base △FGH and a triangular flower bed △PQR. A landscape architect confirms that △FGH ~ △PQR. She measures FG = 9 m, GH = 12 m, FH = 15 m, and the corresponding side PQ = 6 m.

(i) Find the scale factor (ratio of similarity) of △FGH to △PQR.
(ii) Find the lengths of sides QR and PR of the flower bed.
(iii) The architect wants to put a decorative border around the fountain base △FGH. Find the perimeter of △FGH. Also, using the scale factor, find the perimeter of the flower bed △PQR.

OR

(iii) The area of the flower bed △PQR is 24 m². Using the property of similar triangles, find the area of the fountain base △FGH.

Accepted Answer

(i) Since △FGH ~ △PQR, corresponding sides are proportional.

∴ Scale factor = FG/PQ = 9/6 = 3/2

∴ The scale factor of △FGH to △PQR is 3 : 2.        …(1 mark)

(ii) Since △FGH ~ △PQR, we have:

FG/PQ = GH/QR = FH/PR

⟹ 9/6 = 12/QR = 15/PR

Finding QR:
9/6 = 12/QR
⟹ QR = (12 × 6)/9 = 72/9 = 8 m

Finding PR:
9/6 = 15/PR
⟹ PR = (15 × 6)/9 = 90/9 = 10 m

∴ QR = 8 m and PR = 10 m.        …(1 mark)

(iii) [Main Option]

Perimeter of △FGH = FG + GH + FH = 9 + 12 + 15 = 36 m

Since △FGH ~ △PQR, the ratio of their perimeters equals the scale factor:

Perimeter of △FGH / Perimeter of △PQR = FG/PQ = 3/2

⟹ 36 / Perimeter of △PQR = 3/2

⟹ Perimeter of △PQR = (36 × 2)/3 = 24 m

∴ Perimeter of △FGH = 36 m and Perimeter of △PQR = 24 m.        …(2 marks)

[OR Option]

By the property of similar triangles:
Area of △FGH / Area of △PQR = (FG/PQ)²

⟹ Area of △FGH / 24 = (9/6)² = (3/2)² = 9/4

⟹ Area of △FGH = 24 × 9/4 = 216/4 = 54 m²

∴ Area of the fountain base △FGH = 54 m².        …(2 marks)

Question 5

A road safety inspector is checking a triangular warning sign board and its scaled-down model. The sign board (△PQR) and the model (△XYZ) are such that ∠P = ∠X and ∠Q = ∠Y. The inspector measures: PQ = 12 cm, QR = 9 cm, XY = 8 cm.

(i) State the similarity criterion that proves △PQR ~ △XYZ.
(ii) Find the length of YZ.
(iii) If the area of △PQR is 108 cm², find the area of △XYZ.
        OR
    If PR = 15 cm, find XZ.

Accepted Answer

(i) Since ∠P = ∠X and ∠Q = ∠Y (two pairs of corresponding angles are equal),
⟹ by the AA (Angle-Angle) similarity criterion, △PQR ~ △XYZ.
∴ The similarity criterion is AA (Angle-Angle). [1 mark]

(ii) Since △PQR ~ △XYZ, their corresponding sides are proportional.

By the property of similar triangles:
PQ/XY = QR/YZ

Substituting the known values:
12/8 = 9/YZ

⟹ YZ = (9 × 8)/12
⟹ YZ = 72/12
∴ YZ = 6 cm [1 mark]

(iii) [Main option]
By the theorem on areas of similar triangles:
"The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides."

Area(△PQR)/Area(△XYZ) = (PQ/XY)²

108/Area(△XYZ) = (12/8)²

⟹ 108/Area(△XYZ) = (3/2)² = 9/4

⟹ Area(△XYZ) = (108 × 4)/9
⟹ Area(△XYZ) = 432/9
∴ Area(△XYZ) = 48 cm² [2 marks]

OR

[Alternate option]
Since △PQR ~ △XYZ, their corresponding sides are proportional.

By the property of similar triangles:
PQ/XY = PR/XZ

Substituting the known values:
12/8 = 15/XZ

⟹ XZ = (15 × 8)/12
⟹ XZ = 120/12
∴ XZ = 10 cm [2 marks]

Question 6

A city park has a triangular fountain area △PQR and a smaller triangular flower bed △STU inside it. A landscape architect notices that ∠P = ∠S = 65° and ∠Q = ∠T = 55°. The architect records the following measurements:
• PQ = 12 m, QR = 9 m, PR = 15 m
• ST = 8 m

Based on this information, answer the following questions:
(i) Which similarity criterion guarantees that △PQR ∼ △STU? Give one reason. [1 mark]
(ii) Find the length of side TU. [1 mark]
(iii) Find the ratio of the area of △PQR to the area of △STU. [2 marks]
    OR
(iii) The perimeter of △PQR is 36 m. Find the perimeter of △STU. [2 marks]

Accepted Answer

(i)
In △PQR and △STU:
∠P = ∠S = 65° (given)
∠Q = ∠T = 55° (given)
⟹ ∠R = ∠U = 180° − 65° − 55° = 60° (angle sum property)
∴ △PQR ∼ △STU by the AA (Angle-Angle) Similarity Criterion.

(ii)
By the Angle-Angle Similarity Criterion, △PQR ∼ △STU (proved above).
By property of similar triangles, corresponding sides are in proportion:
PQ/ST = QR/TU
⟹ 12/8 = 9/TU
⟹ TU = (9 × 8)/12
⟹ TU = 72/12
∴ TU = 6 m

(iii)
By the property of similar triangles:
Area(△PQR)/Area(△STU) = (PQ/ST)²
⟹ Area(△PQR)/Area(△STU) = (12/8)²
⟹ Area(△PQR)/Area(△STU) = (3/2)²
⟹ Area(△PQR)/Area(△STU) = 9/4
∴ The ratio of the area of △PQR to the area of △STU is 9 : 4.

OR

(iii)
Since △PQR ∼ △STU (by AA criterion, proved in part (i)),
corresponding sides are in proportion:
PQ/ST = Perimeter of △PQR / Perimeter of △STU
⟹ 12/8 = 36 / Perimeter of △STU
⟹ Perimeter of △STU = (36 × 8)/12
⟹ Perimeter of △STU = 288/12
∴ Perimeter of △STU = 24 m

Question 7

In △ABC and △DEF, ∠A = ∠D and ∠B = ∠E. If AB = 6 cm, DE = 4 cm, and the perimeter of △DEF is 24 cm, find the perimeter of △ABC.

Accepted Answer

In △ABC and △DEF,
∠A = ∠D (given)
∠B = ∠E (given)
⟹ ∠C = ∠F (∵ angle sum property of a triangle)

By AA Similarity Criterion, △ABC ∼ △DEF.

By the property of similar triangles, the ratio of corresponding sides equals the ratio of their perimeters.

∴ Perimeter of △ABC / Perimeter of △DEF = AB / DE

Substituting the known values:

Perimeter of △ABC / 24 = 6 / 4

⟹ Perimeter of △ABC = (6 / 4) × 24

⟹ Perimeter of △ABC = 36

∴ The perimeter of △ABC is 36 cm.

Question 8

In the given figure, △ABC is a right triangle, right-angled at B. D is a point on BC such that AD ⊥ BC. If AB = 6 cm, BC = 10 cm, prove that △ABC ~ △ADB and hence find the length of AD.

Accepted Answer

**Diagram:** Right △ABC with ∠B = 90°, D on BC such that AD ⊥ BC (i.e., ∠ADB = 90°).

**To Prove:** △ABC ∼ △ADB

In △ABC and △ADB,

∠ABC = ∠ADB = 90°  ...(each is a right angle)

∠BAC = ∠BAD  ...(common angle at A)

∴ By AA similarity criterion, △ABC ∼ △ADB  ...(i)  **Hence proved.**

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**Finding AD:**

First, find AC using the Pythagoras Theorem in △ABC:

⟹ AC² = AB² + BC²

⟹ AC² = 6² + 10² = 36 + 100 = 136

⟹ AC = √136 = 2√34 cm

Since △ABC ∼ △ADB (from (i)), their corresponding sides are proportional:

$$\frac{AB}{AD} = \frac{BC}{AB}$$

⟹ AD = \frac{AB²}{BC} = \frac{6²}{10} = \frac{36}{10}

∴ **AD = 3.6 cm**

Question 9

In the given figure, ABCD is a trapezium with AB ∥ DC. The diagonals AC and BD intersect at point O. Prove that OA/OC = OB/OD.

Accepted Answer

**Given:** Trapezium ABCD with AB ∥ DC. Diagonals AC and BD intersect at O.

**To Prove:** OA/OC = OB/OD

**Construction:** Draw EF passing through O and parallel to AB (and DC), meeting AD at E and BC at F.

**Proof:**

In △ACD, EO ∥ DC (by construction)

⟹ By the Basic Proportionality Theorem (BPT), a line drawn parallel to one side of a triangle divides the other two sides in the same ratio.

⟹ AO/OC = AE/ED     ...(i)

In △ABD, EO ∥ AB (by construction)

⟹ By the Basic Proportionality Theorem,

⟹ DE/EA = DO/OB

⟹ AE/ED = OB/OD     ...(ii)

From (i) and (ii),

⟹ OA/OC = OB/OD

**Hence proved.**

Question 10

In △ABC and △DEF, it is given that ∠A = ∠D and AB/DE = AC/DF. Using the SAS Similarity Criterion, prove that △ABC ∼ △DEF. Hence, find the value of EF if AB = 4 cm, DE = 6 cm, and BC = 5 cm.

Accepted Answer

Given: In △ABC and △DEF, ∠A = ∠D and AB/DE = AC/DF.

To Prove: △ABC ∼ △DEF.

Construction: Mark point P on DE such that DP = AB, and point Q on DF such that DQ = AC. Join PQ.

Proof:

In △ABC and △DPQ:
  DP = AB          (by construction)
  DQ = AC          (by construction)
  ∠D = ∠A          (given)

⟹ △ABC ≅ △DPQ     (by SAS congruence criterion)

⟹ ∠ABC = ∠DPQ      (CPCT)  ...(i)

Now, since AB/DE = AC/DF
and DP = AB, DQ = AC,
⟹ DP/DE = DQ/DF

⟹ By the converse of the Basic Proportionality Theorem (BPT), PQ ∥ EF.

⟹ ∠DPQ = ∠DEF     (corresponding angles, PQ ∥ EF)  ...(ii)

From (i) and (ii):
  ∠ABC = ∠DEF

In △ABC and △DEF:
  ∠A = ∠D           (given)
  ∠ABC = ∠DEF       (proved above)

⟹ △ABC ∼ △DEF      (by AA Similarity Criterion)

Hence proved.

─────────────────────────────
Finding EF:

Since △ABC ∼ △DEF,
  BC/EF = AB/DE

⟹ 5/EF = 4/6

⟹ EF = (5 × 6)/4

⟹ EF = 30/4

∴ EF = 7.5 cm

Question 11

In the given figure, PQ is a chord of a circle and PT is the tangent at P such that ∠QPT = 60°. Prove that PR · PQ = PS · PT is NOT what is asked here.

Actually: In △ABC, D is a point on side AB and E is a point on side AC such that DE ∥ BC. If AD = 3x − 1, DB = x + 3, AE = 2x + 1 and EC = x + 5, find the value of x.

Accepted Answer

In △ABC, DE ∥ BC.

By the Basic Proportionality Theorem (BPT), if a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.

∴ AD/DB = AE/EC

⟹ (3x − 1)/(x + 3) = (2x + 1)/(x + 5)

⟹ (3x − 1)(x + 5) = (2x + 1)(x + 3)          [cross-multiplying]

⟹ 3x² + 15x − x − 5 = 2x² + 6x + x + 3

⟹ 3x² + 14x − 5 = 2x² + 7x + 3

⟹ 3x² − 2x² + 14x − 7x − 5 − 3 = 0

⟹ x² + 7x − 8 = 0

⟹ x² + 8x − x − 8 = 0

⟹ x(x + 8) − 1(x + 8) = 0

⟹ (x − 1)(x + 8) = 0

⟹ x = 1  or  x = −8

∵ x = −8 gives DB = −8 + 3 = −5, which is negative, and a length cannot be negative, so x = −8 is rejected.

∴ x = 1

Question 12

Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Using the above theorem, if △ABC ∼ △DEF such that BC = 6 cm, EF = 10 cm and ar(△DEF) = 125 cm², find ar(△ABC).

Accepted Answer

**Theorem:** The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

**Given:** △ABC ∼ △PQR

**To Prove:** ar(△ABC) / ar(△PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)²

**Construction:** Draw AM ⊥ BC and PN ⊥ QR.

**Figure:** [Diagram — △ABC with altitude AM from A to BC; △PQR with altitude PN from P to QR; correspondence A↔P, B↔Q, C↔R labelled]

**Proof:**

ar(△ABC) = ½ × BC × AM  …(i)

ar(△PQR) = ½ × QR × PN  …(ii)

Dividing (i) by (ii):

ar(△ABC) / ar(△PQR) = (BC × AM) / (QR × PN)  …(iii)

Now, since △ABC ∼ △PQR, ∠B = ∠Q  (corresponding angles of similar triangles are equal)

In △ABM and △PQN:
∠B = ∠Q  (proved above)
∠AMB = ∠PNQ = 90°  (by construction)

⟹ △ABM ∼ △PQN  (by AA similarity criterion)

⟹ AM/PN = AB/PQ  …(iv)

Also, since △ABC ∼ △PQR:

AB/PQ = BC/QR = AC/PR  …(v)  (corresponding sides of similar triangles are proportional)

Substituting (iv) and (v) into (iii):

ar(△ABC) / ar(△PQR) = (BC/QR) × (AM/PN) = (BC/QR) × (AB/PQ) = (BC/QR) × (BC/QR) = (BC/QR)²

∴ ar(△ABC) / ar(△PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)²

**Hence proved.**

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**Application:**

Given: △ABC ∼ △DEF, BC = 6 cm, EF = 10 cm, ar(△DEF) = 125 cm²

By the theorem just proved:

ar(△ABC) / ar(△DEF) = (BC/EF)²

⟹ ar(△ABC) / 125 = (6/10)²

⟹ ar(△ABC) / 125 = 36/100

⟹ ar(△ABC) = (36/100) × 125

⟹ ar(△ABC) = 36 × 125 / 100 = 4500 / 100

∴ ar(△ABC) = 45 cm²

Question 13

In the given figure, △ABC is a triangle in which DE ∥ BC, where D is on AB and E is on AC. If AD = x cm, DB = (x − 2) cm, AE = (x + 2) cm and EC = (x − 1) cm, find the value of x. Hence, also prove that △ADE ~ △ABC, stating clearly the criterion of similarity used.

Accepted Answer

**Given:** In △ABC, DE ∥ BC, AD = x, DB = (x − 2), AE = (x + 2), EC = (x − 1) (all in cm).

**To find:** Value of x; then prove △ADE ∼ △ABC.

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**Part 1 — Finding x (3 marks)**

Since DE ∥ BC, by the **Basic Proportionality Theorem (Thales' Theorem)**:

If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.

∴ AD/DB = AE/EC

⟹ x/(x − 2) = (x + 2)/(x − 1)

⟹ x(x − 1) = (x + 2)(x − 2)       [Cross-multiplying]

⟹ x² − x = x² − 4

⟹ −x = −4

∴ **x = 4**

Verification of values:
 AD = 4 cm, DB = 2 cm, AE = 6 cm, EC = 3 cm
 AD/DB = 4/2 = 2;  AE/EC = 6/3 = 2  ✓

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**Part 2 — Prove △ADE ∼ △ABC (2 marks)**

From Part 1, DE ∥ BC (given).

In △ADE and △ABC:

**Step 1:** ∠DAE = ∠BAC  ...(i)
(Common angle at vertex A)

**Step 2:** ∠ADE = ∠ABC  ...(ii)
(Corresponding angles, since DE ∥ BC and AB is a transversal)

**Step 3:** By the **AA (Angle–Angle) Similarity Criterion** — if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

From (i) and (ii):

∴ **△ADE ∼ △ABC**  (AA criterion)

Hence proved.

Question 14

Prove that if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side (Converse of Basic Proportionality Theorem). Using this theorem, if in △PQR, a line l intersects PQ at S and PR at T such that PS/SQ = PT/TR, and ST is produced to meet QR extended at U, find ∠STQ if ∠PQR = 65° and ∠PRQ = 50°.

Accepted Answer

**Proof of Converse of Basic Proportionality Theorem (Converse of BPT)**

**Statement:** If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.

**Given:** In △ABC, a line DE intersects AB at D and AC at E such that AD/DB = AE/EC.

**To Prove:** DE ∥ BC

**Construction:** Draw DF ∥ BC, where F lies on AC.

**Proof:**

Since DF ∥ BC (by construction), by the Basic Proportionality Theorem (BPT):

AD/DB = AF/FC ...(i)

But it is given that:

AD/DB = AE/EC ...(ii)

From (i) and (ii):

AF/FC = AE/EC

⟹ AF/FC + 1 = AE/EC + 1

⟹ (AF + FC)/FC = (AE + EC)/EC

⟹ AC/FC = AC/EC

⟹ FC = EC

Since F and E are both points on AC with FC = EC, point F coincides with point E.

∴ DE coincides with DF.

Since DF ∥ BC (by construction), we conclude:

**DE ∥ BC.**

**Hence proved.**

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**Application Part:**

**[Diagram: △PQR with S on PQ and T on PR; ST is a transversal; angles at Q and R marked.]**

In △PQR, it is given that PS/SQ = PT/TR.

∴ By the Converse of the Basic Proportionality Theorem, ST ∥ QR.

Now, since ST ∥ QR, and SQ is a transversal cutting parallel lines ST and QR:

∠STQ + ∠TQR = 180°   (co-interior angles, ST ∥ QR, SQ is transversal)

In △PQR:

∠PQR + ∠PRQ + ∠QPR = 180°   (angle sum property of triangle)

⟹ 65° + 50° + ∠QPR = 180°

⟹ ∠QPR = 180° − 115° = 65°

Now, ∠TQR = ∠PQR = 65°   (same angle, S lies on PQ)

Since ST ∥ QR and SQ is a transversal:

∠STQ + ∠TQR = 180°   (co-interior / consecutive interior angles)

⟹ ∠STQ + 65° = 180°

⟹ ∠STQ = 180° − 65°

∴ **∠STQ = 115°**

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