Question

The length of two adjacent sides of a parallelogram are 17 cm and 12cm. one of its diagonals is 25 cm long. Find the area of the parallelogram. Also find the length of altitude from vertex to the side of length 12 cm. ​

Answer 1

Answer:

Area of triangle =

s(s−a)(s−b)(s−c)

Area of parallelogram=2× Area of triangle formed by the adjacent sides and the diagonal

Sides are 17cm,12cm,25cm

Semi perimeter(s)=

2

12+17+25

=

2

54

=27cm

Area =

27(27−12)(27−17)(27−25)

⇒A=

27⋅15⋅10⋅2

⇒A=

3×3×3×3×5×2×5×2

⇒A=3×3×5×2

⇒A=90cm

2

∴ Area of parallelogram =2(90)=180cm

2

.

h=?; b=12cm

A=b×h

⇒180=12×h

⇒

12

180

=h

∴h=20cm

∴ Length of altitude =h=20cm.

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Answer 2

Question

The length of two adjacent sides of a parallelogram are 17 cm and 12cm. one of its diagonals is 25 cm long. Find the area of the parallelogram. Also find the length of altitude from vertex to the side of length 12 cm

Solution

in△BCD:

Let

• a = 17 cm
• b = 12 cm
• c = 25 cm

So,

• semi-perimeter

→ s = (a + b + c)/2

→ (17 + 12 + 25)/2

→ 54/2

→ 27 cm

By heron's formula

Area of △BCD = $\sf \sqrt{s(s-a)(s-b)(s-c)}$

$\sf→ \sqrt{27(27-17)(27-12)(27-25)}</p><p>\\ \sf→ \sqrt{27(10)(15)(2)}</p><p>\\ \sf→ 90 cm^2$

Now,

Area of parallelogram ABCD

→ 2 x Area of △BCD

→ 2 x 90

→ $\sf 180 \:cm^2$

Also,

• Area of parallelogram ABCD = DC x AE

Therefore,

→ 180 = 12 x AE

→ AE = 180/12 = 15 cm

Hence

• the length of the altitude is 15cm
• Area of parallelogram is $\sf 180\:cm^2$