Question

Two identical triangular pieces of paper are pasted one on top of the other such that they overlap in a square of side 4 cm as shown below.
The resulting shape covers an area of
(with explanation)
A. 136 cm²
B. 120 cm²
C. 104 cm²
D. 88 cm²

Answer 1

Answer:

Here area of this shape = Area of two triangles - Area of square ( as squares area is common here)

Area of 1 triangle = (8+4)*(6+4)/2

12*10/2=60cm².

There fore area of 2nd triangle =60cm² as both are identicle

60cm²+60cm²

120cm²

Area of square =4*4=16cm²

120-16 = 104 cm ²

Therefore C. is correct ✔️✔️

Answer 2

$\huge\underbrace\red{\dag Given:- \dag}$

$\boxed{104 cm^{2}}$

$\huge\underbrace\pink {\dag To Find:- \dag}$

Area of the figure

$\huge\underbrace\green {\dag Formula\:Used:- \dag}$

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

$Area of \square = side \time side$

$\huge\underbrace\green {\dag some \ things \ to \ know :-\dag}$

We have to add 4 to height and base because the square length is 4 cm

• $Height \implies 6+4 \implies 10 cm$

• $Base \implies 8+4 \implies 12 cm$

$\huge\underbrace\purple{\dag Solution:- \dag}$

Area of $2 \triangle \ \ A_1 = 2 (\frac{1}{2} \times base \times height )$

$=2 (\frac{1}{2} \times 12 \times 10)$

$=2 (\frac{1}{2} 120)$

$=\frac{240}{2}$

$=120 cm^{2}$

Hence area of $2 \triangle is 120$

Now let’s find area of overlap (square) $A_2 = side \times side$

$= 4 \times 4$

$=16 cm^{2}$

Hence the area of overlapped square is 16

Now find the result area of the figure $=A_1-A_2$

$=120-16$

$=104 cm^{2}$

So area of the figure is $\boxed{C.104cm^{2}}$

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