Question

please answer this question​

Answer 1

Answer:

mαtє hєrє íѕ ur αnѕ

hσpє ít wíll hєlpѕ чσu

Answer 2

Solution:-

Q.5

For question 5. we're going to use similar triangles.

Let's consider a semi-circle, and choose any point on the perimeter. The triangle formed by diameter and the point is always right-angled.

We choose two triangles below, which are similar by AA postulate.

• $\triangle ACH\sim\triangle CBH\ \mathrm{(\because AA\ Postulate)}$

Now, the corresponding sides are equal in ratio.

$\rightarrow\overline{AH}:\overline{CH}=\overline{CH}:\overline{BH}$

$\rightarrow \overline{CH}^2=\overline{AH}\cdot\overline{BH}$

Thus, this is the relation between the three sides.

Let $\overline{AH}=7.5$, and $\overline{BH}=1$. Then, $\overline{CH}=\sqrt{7.5}$. Hence, we can represent the value geometrically. The steps are as follows.

Q.5 Steps

Draw a line segment $\overline{AH}=7.5$, on to the left of the origin.

Draw a line segment $\overline{BH}=1$, on to the right of the origin.

Let $\overline{AB}$ be the diameter of the circle.

The height $\overline{CH}=\sqrt{7.5}$ starts from the origin. Now there forms a right triangle as the diagram is. Then, all we have to do is to use a compass and mark it on the number line.

So, we're done!

Q.6

For question 6. we're going to use the concepts of exponents. Let's start.

Given value

$=\dfrac{243^{\frac{3}{5} }\times 25^{\frac{3}{2} }}{625^{\frac{1}{2} }\times 8^{\frac{4}{3} }\times 16^{\frac{5}{4} }}$

$=\dfrac{(3^5)^{\frac{3}{5} }\times (5^2)^{\frac{3}{2} }}{(5^4)^{\frac{1}{2} }\times (2^3)^{\frac{4}{3} }\times (2^4)^{\frac{5}{4} }}$

$=\dfrac{3^3\times 5^3}{5^2\times 2^4\times 2^5}$

$=\dfrac{27\times 5}{16\times 32}$

$=\boxed{\dfrac{135}{512} }$.

So, we're done!

Learn More:

In the given diagram,

$\rightarrow \overline{AC}\cdot\overline{BC}=\overline{AB}\cdot\overline{CH}$ [Triangle Area]

$\rightarrow \overline{AC}^2=\overline{AH}\cdot\overline{AB}$ [Similarity]

$\rightarrow \overline{BC}^2=\overline{BH}\cdot\overline{AB}$ [Similarity]