The height of an equilateral triangle is 10 cm. find its area.

Answer: Option C

Explanation:

\begin{aligned}
\text{Area of triangle, A1 = }\frac{1}{2}*base*height \\
= \frac{1}{2}*15*12 = 90 cm^2 \\
\text{Area of second triangle =} 2*A1 \\
= 180 cm^2 \\
\frac{1}{2}*20*height = 180 \\
=> height = 18 cm


\end{aligned}

Answer: Option C

Explanation:

\begin{aligned}
\frac{a^2}{b^2} = \frac{225}{256} \\
\frac{15}{16} \\
<=> \frac{4a}{4b} = \frac{4*15}{4*16} \\
= \frac{15}{16} = 15:16
\end{aligned}

Answer: Option B

Explanation:

We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cm

Area of third Square = 10*10 - 8*8
= 36 cm

So side of third square = 6 [because area of square = side*side]
Perimeter = 4*Side = 4*6 = 24 cm

Answer: Option A

Explanation:

Clearly first we need to find the areas of the given squares, for that we need its side.

Side of sqaure = Perimeter/4

So sides are,

\begin{aligned}
\left(\frac{24}{4}\right),\left(\frac{32}{4}\right),\left(\frac{40}{4}\right),\left(\frac{76}{4}\right),\left(\frac{80}{4}\right) \\
= 6,8,10,19,20 \\
\text{Area of new square will be }\\
= [6^2+8^2+10^2+19^2+20^2] \\
= 36+64+100+361+400 \\
= 961 cm^2 \\
\text{Side of new Sqaure =}\sqrt{961} \\
= 31 cm \\
\text{Required perimeter =}(4\times31) \\
= 124 cm

\end{aligned}

Answer: Option A

Explanation:

\begin{aligned}
2\pi R1 = 4 \pi \\
=> R1 = 2 \\
2\pi R2 = 8 \pi \\
=> R2 = 4 \\
\text{Original Area =} 4\pi * 2^2 \\
= 16 \pi \\
\text{New Area =} 4\pi * 4^2 \\
= 64 \pi
\end{aligned}

So the area quadruples.