Question

The area of a parallelogram ABCD in which AB = 12 cm, BC = 9 cm and diagonal AC = 15 cm is k cm2. Find the value of k-100/4

Answer 1

Given:-

• ABCD is a //gm.
• AB = 12cm
• BC = 9cm
• AC = 15cm
• Area of //gm :- k cm²

To find:-

• $\sf{k - \dfrac{100}{4}}$

Solution:-

Area of ∆ = $\sf{ \dfrac{1}{2} \times b \times h}$

$\implies \sf{ A = \dfrac{1}{2} \times 12 \times h}$ ------(1)

★ Also find area of ∆ using herons formula:-

• $\sf{ \sqrt{ s(s - a)(s - b)(s - c)}}$

★ Firstly finding semiperimeter(s) = ?

• $\sf{ \dfrac{ a + b + c}{2}}$

• $\sf{ \dfrac{ 12 + 9 + 15}{2}}$

• $\sf{ \cancel{ \dfrac{36}{2}}}$

• $\sf{ 18cm}$

★ Put value of S in formula:-

• $\sf{ \sqrt{ 18(18 - 12)(18 - 15)(18 - 9)}}$

• $\sf{ \sqrt{ 18(6)(3)(9)}}$

• $\sf{ \sqrt{3 \times 3 \times 2 \times 3 \times 2 \times 3 \times 3 \times 3}}$

• $\sf{ 3 \times 3 \times 3 \times 2}$

• $\sf{ 54 cm^2}$

$\implies$ So, area of ∆ ABC is 54cm².

★ Put the area in eq(1):-

$\implies \sf{ 54 = \dfrac{1}{2} \times 12 \times h}$

$\implies \sf{ 54 = 6 \times h}$

$\implies \sf{ \cancel{ \dfrac{54}{6}} = h}$

$\implies \sf{h = 9cm}$

So,

Area of //gm :- Base × Height

$\implies \sf{ A = 12 \times 9}$

$\implies \sf{ A = 108cm^2}$

Also, Area of //gm :- k

$\implies \sf{ k = 108cm^2}$

Therefore,

$\implies \sf{ k - \dfrac{100}{4}}$

★ Putting value of k :-

$\implies \sf{ 108 - \dfrac{100}{4}}$

$\implies \sf{ \dfrac{432 - 100}{4}}$

$\implies \sf{ \dfrac{332}{4}}$

$\implies \sf{83}$

_______________________

Answer 2

$\huge\purple{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{Value \ of \ k-\frac{100}{4} \ is \ 83.}$

$\sf\orange{Given:}$

$\sf{In \ a \ parallelogram \ ABCD,}$

$\sf{\implies{AB=12 \ cm,}}$

$\sf{\implies{BC=9 \ cm,}}$

$\sf{\implies{Diagonal \ AC=15 \ cm}}$

$\sf{\implies{Area \ of \ parallelogram \ is \ k \ cm^{2}}}$

$\sf\pink{To \ find:}$

$\sf{The \ value \ of \ k-\frac{100}{4}}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{In, \ \triangle \ ABC,}$

$\sf{AB=12 \ cm,}$

$\sf{BC=9 \ cm,}$

$\sf{AC=15 \ cm.}$

$\sf{By \ herons \ formula}$

$\sf{s=\frac{a+b+c}{2}}$

$\sf{s=\frac{12+9+15}{2}}$

$\sf{s=\frac{36}{2}}$

$\sf{\therefore{s=18}}$

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

$\sf{=\sqrt{18(18-12)(18-9)(18-15)}}$

$\sf{=\sqrt{18\times6\times9\times3}}$

$\sf{=\sqrt{3\times3\times2\times3\times2\times3\times3\times3}}$

$\sf{=3\times3\times3\times2}$

$\sf{=54 \ cm^{2}}$

$\sf{Also \ area \ of \ \triangle \ ABC=\frac{1}{2}\times \ base\times \ height}$

$\sf{Now, \ let \ AB \ be \ base.}$

$\sf{\therefore{Area \ of \ \triangle \ ABC=\frac{1}{2}\times \ AB\times \ h}}$

$\sf{54=\frac{1}{2}\times12\times \ h}$

$\sf{h\times6=54}$

$\sf{h=\frac{54}{6}}$

$\sf{\therefore{h=9 \ cm}}$

$\sf{Height \ is \ 9 \ cm}$

_________________________________

$\sf{Area \ of \ parallelogram=Base\times \ height}$

$\sf{...formula}$

$\sf{\therefore{Area \ of \ parallelogram \ ABCD=12\times9}}$

$\sf{=108 \ cm^{2}}$

$\sf{\therefore{k=108}}$

$\sf{\implies{k-\frac{100}{4}=108-\frac{100}{4}}}$

$\sf{=108-25}$

$\sf{=83}$

$\sf\purple{\tt{\therefore{Value \ of \ k-\frac{100}{4} \ is \ 83.}}}$