Question

In the adjoining figure, ABC is a triangle in which ∠B = 90°. M and N are the mid-points of sides AB and A, respectively. If AB = 5 cm, BC = 12 cm, find

1. Perimeter of ◻MNCB
2. Area of ◻MNCB

Answer 1

Given:

Triangle ABC with ∠B = 90°, AB = 5 cm, BC = 12 cm.

M and N are midpoints of AB and AC respectively.

By the midpoint (mid‑segment) theorem,

MN || BC and MN = `1/2` × BC.

Step-wise calculation:

1. Find AC (hypotenuse) using Pythagoras:

`AC = sqrt(AB^2 + BC^2)`

= `sqrt(5^2 + 12^2)`

= `sqrt(25 + 144)`

= `sqrt(169)`

= 13 cm

2. Place coordinates for convenience: 

B(0, 0), C(12, 0), A(0, 5). 

Then M, midpoint of AB = (0, 2.5).

N, midpoint of AC = `((0 + 12)/2, (5 + 0)/2)` = (6, 2.5).

3. Side lengths of quadrilateral M–N–C–B:

MN = Distance between (0, 2.5) and (6, 2.5) = 6 cm also = `1/2` × BC = 6.

NC = Distance between (6, 2.5) and (12, 0)

= `sqrt(6^2 + 2.5^2)`

= `sqrt(36 + 6.25)`

= `sqrt(42.25)`

= 6.5 cm

CB = BC = 12 cm.

BM = Distance between (0, 0) and (0, 2.5) = 2.5 cm.

4. Perimeter of MNCB: 

Perimeter = MN + NC + CB + BM

= 6 + 6.5 + 12 + 2.5

= 27 cm

5. Area of MNCB: 

MNCB is a trapezoid with parallel bases BC = 12 and MN = 6.

And Height = Distance between the parallel lines = 2.5.

Area = `1/2` × (sum of bases) × height

= `1/2 xx (12 + 6) xx 2.5`

= 0.5 × 18 × 2.5

= 22.5 cm²

Perimeter of quadrilateral MNCB = 27 cm.

Area of quadrilateral MNCB = 22.5 cm².