Question

A triangle and a parallelogram have the same base and the same area if the sides of the triangle are 26cm 28cm and 30 cm and the parallelogram stands on the base 20 cm find the height of the parallelogram?

help me out anyone.​

Answer 1

${\bold{\huge{\underline{QUESTION-}}}}$

A triangle and a parallelogram have the same base and the same area if the sides of the triangle are 26cm 28cm and 30cm and the parallelogram stands on the base 20 cm find the height of the parallelogram?

${\bold{\huge{\underline{SOLUTION-}}}}$

Given that the three sides of triangle are 26cm 28cm and 30cm.

Given

a = 26cm

b = 28cm

c = 30cm

Base of the parallelogram = 20cm

As the semi-perimeter is the half of the sum of sides of triangle.

$s \: = \: \frac{a + b + c}{2} \\ \\ s \: = \: \frac{26 + 28 + 30}{2} \\ \\ s \: = \: \frac{84}{2} \\ \\ s \: = \: 42cm$

Therefore area of triangle =

$= \: \sqrt{s \: ( \: s \: - \: a \: ) \: ( \: s \: - \: b \: ) \: ( \: s \: - \: c \: )} \\ \\ = \: \sqrt{42 \: ( \: 42 \: - \: 26 \: ) \: ( \: 42 \: - \: 28 \: ) \: ( \: 42 \: - \: 30 \: )} \\ \\ = \: \sqrt{42 \times 16 \times 14 \times 12} \\ \\ = \: \sqrt{7 \times 6 \times 4 \times 4 \times 7 \times 2 \times 2 \times 6} \\ \\ = \: \sqrt{ {(7)}^{2} \times {(6)}^{2} \times {(4)}^{2} \times {(2)}^{2} } \\ \\ = \: \sqrt{ {(7 \times 6 \times 4 \times 2)}^{2} } \\ \\ = \: \sqrt{ {(336)}^{2} } \\ \\ = \: 336 {cm} ^{2}$

But,

Area of parallelogram = Area of Δ ( given )

$= \: base \: \times \: height \: = \: 336 \\ = \: 28 \: \times \: height \: = \: 336 \\ = \: height \: \times \: \frac{336}{28} \\ = \: 12cm$

Answer is ( H ) Height = 12cm

Answer 2

Answer:

The height of the parallelogram is 12 cm.

Step-by-step explanation:

Given Data -

x = 26cm.

y = 28cm.

z = 30cm.

Base of the parallelogram = 20 cm

_____{Sides are denoted by x , y & z}

➟ First Step -

Find the area of a triangle by heron’s formula and area of parallelogram.

➟ Second Step -

Equate their areas to calculate the height of a parallelogram.

Process -

Let s be the semi perimeter of the triangle.

$\sf \implies s = \dfrac{(a+b+c)}{2}$

$\sf \implies s=\dfrac{(26+28+30)}{2}$

$\sf \implies s= \dfrac{84}{2}= 42 cm$

$\bf{\implies s = 42 cm}$

As we know -

$\bigstar\;\boxed{\sf Area\;of\;the\;triangle = \sqrt{s (s-a) (s-b) (s-c)}}$

____________{ Heron's Formula }

Values in Equation,

$\sf\\ \\\implies \sqrt{42(42-26)(42-28)(42-30)}$

$\sf\\ \\ \implies \sqrt{42 \times 16 \times 14 \times12}$

$\sf\\ \\\implies \sqrt{7 \times 6 \times 4 \times 4 \times 7 \times 2 \times 2 \times 6}$

$\sf\\ \\\implies \sqrt{(7)^2 \times (6)^2 \times (4)^2\times (2)^2}$

$\sf\\ \\ \implies \sqrt{7 \times 6 \times 4 \times 2^2}$

$\sf\\ \\ \implies \sqrt{336^2}$

$\bf\\ \\ \implies 336cm^2$

$\rule{300}{1.5}$

Let height of parallelogram be h.

Base = 28

________{ Given }

As given,

Area of parallelogram = Area of triangle.

As we know,

$\bigstar\;\boxed{\sf Area\;of\;Parallelogram=Base\times Height}$

___________________

Values in Equation,

$\sf\\ \\\implies 28\times H = 336\\ \\ \implies H = \dfrac{336}{28}cm\\ \\\implies H = 12 cm$

Therefore, The height of the parallelogram is 12 cm.