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Syllabus

$8.\mathrm{In}\mathrm{the}\mathrm{adjoining}\mathrm{figure},\angle \mathrm{BCD}=\angle \mathrm{ADC}\mathrm{and}\angle \mathrm{BCA}=\angle \mathrm{ADB}.\mathrm{Show}\mathrm{that}.\phantom{\rule{0ex}{0ex}}\left(\mathrm{i}\right)\u25b3\mathrm{ACD}\cong \u25b3\mathrm{BDC}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\mathrm{BC}=\mathrm{AD}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\angle \mathrm{A}=\angle \mathrm{B}\phantom{\rule{0ex}{0ex}}$

angleAFE =137degree

Find the value of x and y (in the figure)

Q. In the alongside diagram, the bisectors of interior angles of the parallelogram PQRS enclose a quadrilateral ABCD.

Show that :

(i) $\angle $PSB+$\angle $SPB = 90$\xb0$ (ii) $\angle $PBS = 90$\xb0$ (iii) $\angle $ABC = 90$\xb0$ (iv) $\angle $ADC = 90$\xb0$

(v) $\angle $A = 90$\xb0$ (vi) ABCD is a rectangle

BG=21m,

AM=18m

DN=12.8m, and

ED=14m

and the distance between CB and DA

With step by step procedure. How to do ??

Chapter - qaudrilaterals

$\mathbf{10}\mathbf{.}\mathrm{In}\mathrm{the}\mathrm{given}\mathrm{figure},\mathrm{ABCDE}\mathrm{is}\mathrm{a}\mathrm{pentagon}\mathrm{inscribed}\mathrm{in}\mathrm{a}\mathrm{circle}.\mathrm{If}\mathrm{AB}=\mathrm{BC}=\mathrm{CD},\angle \mathrm{BCD}=110\xb0\mathrm{and}\angle \mathrm{BAE}=120\xb0,\mathrm{find}:\phantom{\rule{0ex}{0ex}}\left(\mathrm{i}\right)\angle \mathrm{ABC}\left(\mathrm{ii}\right)\angle \mathrm{CDE}\left(\mathrm{iii}\right)\angle \mathrm{AED}\left(\mathrm{iv}\right)\angle \mathrm{EAD}$

Find it

Q. ABCD is a rectangle. If $\angle $BPC = 124$\xb0$. Calculate : (i) $\angle $BAP (ii) $\angle $ADP

Angle C = (3x - 10). Find each angle of trapezium.

Q.4. Is it possible to construct a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 5.5 cm, DA = 6 cm and BD = 9 cm? If not, give reason.

and ?C=135?

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