In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are A, B, C, as marked. Class 9

In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are A, B, C, as marked. Class 9

Question 1.

In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are A, B, C, as marked. Show that the area of the rectangle is $\frac{2(A+C)(B+C)}{C}$ Class 9

Solution:

Given:

Three triangles inside a rectangle with areas A, B, C

To prove:

Area of rectangle = $\frac{2(A+C)(B+C)}{C}$

Proof:

Let PQRS be a quadrilateral and A, B and C are regions represented by ∆PML, ∆PMN and ∆LMN respectively.

Let w be base and h be height of rectangle PQRS.

∴ PQ = SR = w and SP = QR = h

Let ∆LMN (Region C) has base MN = y and height RN = x

Using Area of triangle = $\frac{1}{2}$ × base × height

∴ A = Area of ∆PML = $\frac{1}{2}$ × x ×(w - y) ... (i)

∴ B = Area of ∆PMN = $\frac{1}{2}$ × y × (h - x) ...(ii)

∴ C = Area of ∆LMN = $\frac{1}{2}$ × x × y ...(iii)

Now,

Adding (i), (ii) we get

A + C = $\frac{1}{2}$ × x × (w - y) + $\frac{1}{2}$ × x × y

= $\frac{1}{2}$ (wx -xy + xy)

A + C = $\frac{1}{2}$wx ........(iv)

Adding (ii) and (iii), we get

B + C = $\frac{1}{2}$ × y × (h - x) + $\frac{1}{2}$ xy

= $\frac{1}{2}$ (yh - xy + xy)

B + C = $\frac{1}{2}$ × yh ...(v)

$\begin{aligned} \text { R.H.S. } & =\frac{2(A+C)(B+C)}{C} \\ & =\frac{2 \times \frac{1}{2} w x \times \frac{1}{2} y h}{\frac{1}{2} x y}\end{aligned}$

...[From (iii), (iv) and (v)]

= w × h

= Area of rectangle PQRS

= L.H.S.

∴ Area of rectangle = $\frac{2(A+C)(B+C)}{C}$

[Note: Question has been modified,]