Maths test

Q1.

The three sides of a triangle are 5m, 6m and 7m respectively, and then what is the area of the given triangle. Ans.

Three sides of the triangle are $5m, 6m$ & $7m$.

$∴s = (5 + 6 + 7)/2 = 9m$

$Area = \sqrt{s(s-a)(s-b)(s-c)}$

$\implies \sqrt{9(9-5)(9-6)(9-7)}$

$\implies \sqrt{9 X 4 X 3 X 2} = 6\sqrt{6}m^2$

Q2.

In an isosceles right-angled triangle, the perimeter is $20$ meters. Find its area.

Ans.

In an isosceles right angled triangle, $Base = Height$. Let $a$ be the $base$ and $b$ be the $hypotenuse$.

$∴ a + a + b = 20 \implies 2a + b = 20. \newline Also \space b^2 = a^2 + a^2 \implies b^2 = 2a^2$

$\therefore b = \sqrt{2}a.$

$So \space 2a+ \sqrt{2}a = 20 \implies 3.41a = 20 \newline ∴ a = 5.86m$

$Required\space area = (1/2)XaXa = (1/2)X5.86X5.86=17.16m^2$

Q3. Solve for z

$5(z + 1) = 3(z + 2) + 11$

Ans.

$\begin{aligned} 5(z + 1) &= 3(z + 2) + 11\\ 5z + 5 &= 3z + 6 + 11\\ 5z + 5 &= 3z + 17 \\ 5z &= 3z + 17 – 5\\ 5z – 3z &= 12\\ 2z &= 12\\ \therefore z &= 6\\ \end{aligned}$

Q4.

The length of the minute hand of a clock is $14cm$. Find the area swept by the minute hand in $5mins$.

Ans.

Length of minute hand $=$ radius of the clock (circle)

$\therefore$ Radius $(r)$ of the circle $= 14cm$ (given)

Angle swept by minute hand in $60 mins = 360°$

So, the angle swept by the minute hand in $5 mins = 360° × 5/60 = 30°$

We know,

Area of a sector $= (θ/360°) × πr^2$

Now, area of the sector making an angle of $30°$

Q5.

If $x + \dfrac{1}{x} = 99$, find the value of $\dfrac{100x}{2x^2 + 102x + 2}$

Ans.

$\begin{aligned} x + \dfrac{1}{x} &= 99\\ x + \dfrac{1}{x} + 1 &= 99 + 1\\ x + \dfrac{1}{x} - 100 &= - 1\\ \dfrac{x^2 + 1 - 100x}{x} &= - 1\\ x^2 + 1 - 100x &= - x\\ – x^2 – 1 – 1 &= – 100x \\ 100x &= x2 + x + 1 \\ \dfrac{100x}{2x^2 + 102x + 2} &= \dfrac{100x}{2 (x^2 + 51x + 1)} \\ \end{aligned}$

$\implies \dfrac{100x}{2 (x^2 + 50x + x + 1)} \\ \implies \dfrac{100x}{2 (x^2 + x + 1 + 50x)} \\ \implies \dfrac{100x}{2 (100x + 50x)} \\ \implies \dfrac{100x}{2 (150x)} \space\space since\space x2 + x + 1 = 100x \\ \therefore \dfrac{100x}{300x} = \dfrac{1}{3}$

Refrences

• https://www.symbolab.com/
• https://www.mathcha.io/editor