# In Fig. 9.21, if PR = 12 cm, QR = 6 cm and PL = 8 cm, then QM is

a. 6 cm

b. 9 cm

c. 4 cm

d. 2 cm

**Solution:**

It is given that

PR = 12 cm

QR = 6 cm

PL = 8 cm

In __right angled triangle__ PLR

Using the __Pythagoras theorem__

PR² = PL² + LR²

LR² = PR² - PL²

Substituting the values

LR² = 144 - 64

LR² = 80

LR = √80 = 4√5 cm

Here

LR = LQ + QR

LQ = LR - QR

Substituting the values

LQ = 4√5 - 6 cm

__Area of triangle__ PLR

A1 = 1/2 × LR × PL

Substituting the values

A1 = 1/2 × 4√5 × 8 = 16√5 cm²

Area of triangle PLQ

A2 = 1/2 × LQ × PL

Substituting the values

A2 = 1/2 × (4√5 - 6) × 8

A2 = 16√5 - 24 cm²

We know that

Area of triangle PLR = Area of triangle PLQ + Area of triangle PQR

Substituting the values

16√5 = 16√5 - 24 + Area of triangle PQR

Area of triangle PQR = 24 cm²

1/2 × PR × QM = 24

1/2 × 12 × QM = 24

QM = 4 cm

Therefore, QM is 4 cm.

**✦ Try This: **144 unit squares are joined to form a rectangle with the least perimeter. Perimeter of the rectangle is

**☛ Also Check: **NCERT Solutions for Class 7 Maths Chapter 11

**NCERT Exemplar Class 7 Maths Chapter 9 Problem 16**

## In Fig. 9.21, if PR = 12 cm, QR = 6 cm and PL = 8 cm, then QM is a. 6 cm, b. 9 cm, c. 4 cm, d. 2 cm

**Summary:**

In Fig. 9.21, if PR = 12 cm, QR = 6 cm and PL = 8 cm, then QM is 4 cm

**☛ Related Questions:**

- In reference to a circle the value of π is equal to a. area/circumference, b. area/diameter, c. circ . . . .
- Circumference of a circle is always a. more than three times of its diameter, b. three times of its . . . .
- Area of triangle PQR is 100 cm² (Fig. 9.20). If altitude QT is 10 cm, then its base PR is a. 20 cm, . . . .

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