Question
Complete the following activity to solve the quadratic equation
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Similar questions
| Draw a circle and take any point P on the circle. Draw ray OP |
| ↓ |
| Draw perpendicular to ray OP from point P |
Find distance between point $Q(3,-7)$ and point $R(3,3)$
Solution: Suppose $Q\left(x_1, y_1\right)$ and point $R\left(x_2, y_2\right)$
$x_1=3, y_1=-7 \text { and } x_2=3, y_2=3$
Using distance formula,
$ d(Q, R)=\sqrt{\square}$
$\therefore d(Q, R)=\sqrt{\square-100}$
$\therefore d(Q, R)=\sqrt{\square}$
$\therefore d(Q, R)=\square $
→Solution: Suppose $Q\left(x_1, y_1\right)$ and point $R\left(x_2, y_2\right)$
$x_1=3, y_1=-7 \text { and } x_2=3, y_2=3$
Using distance formula,
$ d(Q, R)=\sqrt{\square}$
$\therefore d(Q, R)=\sqrt{\square-100}$
$\therefore d(Q, R)=\sqrt{\square}$
$\therefore d(Q, R)=\square $
A die is rolled. Complete the following activity to find the probability of getting a prime number on the upper face of the die :
A die is rolled. S is the sample space.
A die is rolled. S is the sample space.
A ladder $10\ m$ long reaches a window $8\ m$ above the ground. Find the distance of the foot of the ladder from the base of wall. Complete the given activity.
Activity: As shown in figure suppose
PR is the length of ladder $=10 m$
At P - window, At Q - base of wall, At R - foot of ladder
$\therefore PQ =8 m$
$\therefore Q R=?$
In $\triangle P Q R, m \angle P Q R=90^{\circ}$
By Pythagoras Theorem,
$\therefore PQ ^2+\square= PR ^2$
Here, $P R=10, P Q=\square$
From equation (l)
$ 8^2+Q^2=10^2$
$Q R^2=10^2-8^2$
$Q R^2=100-64$
$Q R^2=\square$
$Q R=6 $
$\therefore$ The distance of foot of the ladder from the base of wall is $6 m$.
→Activity: As shown in figure suppose
PR is the length of ladder $=10 m$
At P - window, At Q - base of wall, At R - foot of ladder
$\therefore PQ =8 m$
$\therefore Q R=?$
In $\triangle P Q R, m \angle P Q R=90^{\circ}$
By Pythagoras Theorem,
$\therefore PQ ^2+\square= PR ^2$
Here, $P R=10, P Q=\square$
From equation (l)
$ 8^2+Q^2=10^2$
$Q R^2=10^2-8^2$
$Q R^2=100-64$
$Q R^2=\square$
$Q R=6 $
$\therefore$ The distance of foot of the ladder from the base of wall is $6 m$.
To draw tangents to the circle from the endpoints of the diameter of the circle.
| Construct a circle with center O. Draw any diameter AB of it |
| ↓ |
| Draw ray OA and ray OB |
| ↓ |
| Construct perpendicular to ray OA from point A |
| ↓ |
| Construct perpendicular to Ray OB from point B |
The first term of an A.P. is 5 and the common difference is 4 . Complete the following activity to find the sum of first 12 terms of this A.P.
In ∆ABC, ∠C is an acute angle, seg AD ⊥ seg BC. Prove that:
AB² = BC² + AC² - 2BC×DC
AB² = BC² + AC² - 2BC×DC