Question

In || gm ABCD, M is a point on AB, ∠DMC = 90°. DM = 15 cm, DC = 17 cm. ∴  Area of || gm ABCD is:

Options

• 60 cm²

• 68 cm²

• 120 cm² 

• 136 cm²

Answer 1

120 cm² 

Explanation:

You have a right-angled triangle ΔDMC inside the parallelogram.

We know two sides: DM = 15 cm and DC (hypotenuse) = 17 cm.

Using the Pythagorean theorem (a² + b² = c²), we can find the third side, MC.

15² + MC² = 172

225 + MC² = 289

MC2 = 289 − 225 = 64

MC = `sqrt64` = 8 cm

Calculate the area of the triangle ΔDMC:

Since ΔDMC is a right-angled triangle, its area is `1/2` × base × height.

Using DM as height and MC as base: Area = `1/2 xx 8 xx 15 = 60` cm²

Relate the triangle’s area to the parallelogram’s area to find the height (h):

The area of a parallelogram is given by: Area = base × height.

In ΔDMC, if we consider DC (17 cm) as the base, then the height corresponding to this base is the perpendicular distance from M to DC. This perpendicular distance is also the height (h) of the parallelogram (because M is on AB, and AB is parallel to DC).

So, we can also write the area of ΔDMC as `1/2 xx DC xx h`

We know Area of ΔDMC = 60 cm² and DC = 17 cm

`60 xx 1/2 xx 17 xx h`

120 = 17 × h

h = `120/17` cm

Calculate the area of the parallelogram:

Now we have the base of the parallelogram (DC = 17 cm) and its height (h = `120/17` cm)

Area of parallelogram ABCD = base × height = `17 xx 120/17`

Area of parallelogram ABCD = 120 cm²