Question
prove the theorem Opposite angles of a cyclic quadrilateral are supplementry.
✓
Answer
self
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In fig., $PM = 10 cm, A(\triangle PQS) = 100$ sq.cm, $A(\triangle QRS) = 110$ sq.cm, then $NR?$
$\triangle P Q S$ and $\triangle Q R S$ having seg $Q S$ common base.
Areas of two triangles whose base is common are in proportion of their corresponding
$ \frac{ A ( PQS )}{ A ( QRS )}=\frac{[}{ NR }$
$\frac{100}{110}=\frac{[}{ NR },$
$NR =[\ldots] cm $
→$\triangle P Q S$ and $\triangle Q R S$ having seg $Q S$ common base.
Areas of two triangles whose base is common are in proportion of their corresponding
$ \frac{ A ( PQS )}{ A ( QRS )}=\frac{[}{ NR }$
$\frac{100}{110}=\frac{[}{ NR },$
$NR =[\ldots] cm $
To draw the graph of $4 x+5 y=19$, complete the following activity to find $y$, when $x=1$.
→→In the adjoining figure, AB || CD || EF. If AC = 5.4, CE = 9, BD = 7.5, then find DF.
$AB \| CD \| EF$
$\therefore \frac{ AC }{ ⬜ }=\frac{ ⬜ }{ DF }$
$\therefore \quad \frac{5.4}{9}=\frac{⬜}{ DF }$
$\therefore \quad DF =\frac{7.5 \times 9}{5.4}$
∴ DF = ⬜
[Given]
⬜
→$AB \| CD \| EF$
$\therefore \frac{ AC }{ ⬜ }=\frac{ ⬜ }{ DF }$
$\therefore \quad \frac{5.4}{9}=\frac{⬜}{ DF }$
$\therefore \quad DF =\frac{7.5 \times 9}{5.4}$
∴ DF = ⬜
[Given]
⬜
Complete the following activity to solve the simultaneous equations
From given figure, In ∆ABC, If AC = 12 cm. then AB =?
Activity: From given figure, In $\triangle ABC , \angle ABC =90^{\circ}, \angle ACB =30^{\circ}$
$\therefore \angle BAC =\square$
$\therefore \triangle ABC$ is $30^{\circ}-60^{\circ}-90^{\circ}$ triangle
$\therefore$ In $\triangle ABC$ by property of $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$ \therefore A B=\frac{1}{2} A C \text { and } \square=\frac{\sqrt{3}}{2} A C$
$\therefore \square=\frac{1}{2} \times 12 \text { and } B C=\frac{\sqrt{3}}{2} \times 12$
$\therefore \square=6 \text { and } B C=6 \sqrt{3} $
→Activity: From given figure, In $\triangle ABC , \angle ABC =90^{\circ}, \angle ACB =30^{\circ}$
$\therefore \angle BAC =\square$
$\therefore \triangle ABC$ is $30^{\circ}-60^{\circ}-90^{\circ}$ triangle
$\therefore$ In $\triangle ABC$ by property of $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$ \therefore A B=\frac{1}{2} A C \text { and } \square=\frac{\sqrt{3}}{2} A C$
$\therefore \square=\frac{1}{2} \times 12 \text { and } B C=\frac{\sqrt{3}}{2} \times 12$
$\therefore \square=6 \text { and } B C=6 \sqrt{3} $
Shri Shantilal purchased 150 shares of FV ₹ 100, for MV ₹ 120. Company paid dividend 7% Complete the following activity to find the rate of return on his investment.
FV =\%100; Number of shares =? 150 ; MV = mathbb * 120
FV =\%100; Number of shares =? 150 ; MV = mathbb * 120
Two coins are tossed simultaneously. Complete the following activity of writing the sample space $(S)$ and expected outcomes of the events :
→To check the rule for the terms of the sequence look at the arrangements and fill the empty boxes suitably.
3,3,3,3,…
3,3,3,3,…
A card is drawn from a well-shuffled pack of 52 playing cards. Complete the following activity to find the probability of the event that the card drawn is a red card:
Suppose S is the sample space. .. n(S) = 52.
Event A Card drawn is a red card.
Suppose S is the sample space. .. n(S) = 52.
Event A Card drawn is a red card.