Question

7. The lengths of the base and the height of a triangle are in the ratio4:5. If the area of the triangle is 40 sq m, find the lengths of itsbase and height.
[Hint: Let base be 4x m and the height be 5x m.
1/2 x 4x x 5x = 40.]
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Answer 1

Answer:

Let base of ∆ = 4x m

and height of ∆ = 5x m

area of ∆ =40 m2

∵ (1/2) base × height = area of ∆

(1/2) × 4x × 5x = 40

10x^2 = 40

x^2 = 4

x = 2

∴ base = 4x = 4 × 2 = 8 m

height = 5x = 5 × 2 = 10 m

∴ 8 m ; 10 m

Answer 2

Question:

The lengths of the base and the height of a triangle are in the ratio4:5. If the area of the triangle is 40 sq m, find the lengths of itsbase and height.

To find:

• Length of base

• Height of triangle

Given :

• The lengths of the base and the height of a triangle are in the ratio of 4:5

• Area of triangle = 40 m²

Let:

• Length of base = 4x

• The height of triangle = 5x

Solution :

We know:

$\checkmark \boxed{ \rm{}Area \: of \: triangle = \frac{1}{2} \times base \times height}$

By using this formula we can find value of x

$: \implies\sf{}Area \: of \: triangle = \dfrac{1}{2} \times base \times height$

put value of area and of base and height which we have supposed

$: \implies\sf{}40 = \dfrac{1}{2} \times 4x \times 5x$

$: \implies\sf{}40 = \dfrac{4x \times 5x}{2}$

$: \implies\sf{}40 = \dfrac{20 \: {x}^{2} }{2}$

$: \implies\sf{}40 \times 2= 20 \: {x}^{2}$

$: \implies\sf{}80= 20 \: {x}^{2}$

$: \implies\sf{} \dfrac{80}{20} ={x}^{2}$

$: \implies\sf{} \dfrac{8\cancel0}{2\cancel0} ={x}^{2}$

$: \implies\sf{} \dfrac{8}{2} ={x}^{2}$

$: \implies\sf{}\cancel \dfrac{8}{2} ={x}^{2}$

$: \implies\sf{}{x}^{2} = 4$

$: \implies\sf{}x = \sqrt{4}$

$: \implies\sf{}x = \sqrt{2 \times 2}$

$: \implies \underline{ \boxed{\sf{}x = 2 \: m}}$

Verification:

$: \implies\sf{}40 = \dfrac{1}{2} \times 4x \times 5x$

$: \implies\sf{}40 = \dfrac{4x \times 5x}{2}$

put value of x in this equation

$: \implies\sf{}40 = \dfrac{4 \times 2\times 5 \times 2}{2}$

$: \implies\sf{}40 = \dfrac{4 \times 2\times 5 \times \cancel2}{ \cancel2}$

$: \implies \underline{ \boxed{\sf{}40 = 40}}$

LHS=RHS

Hence verified!

Now Let's find Length of base and Height of triangle

• Length of base = 4x
• Length of base = 4×2
• Length of base = $\boxed{\bf 8m}$

• Height of triangle = 5x
• Height of triangle = 5× 2
• Height of triangle = $\boxed{\bf 10m}$