Question

In a rectangle $ABCD$ where $AB = 6$ , $BC = 3$ , point $P$ is chosen on $AB $ such that $\angle APD = 2 \angle CPB$. Find $AP$.

score 903 · Answer 1 · 2026-04-16 10:01:19

903

Views

In a rectangle $ABCD$ where $AB = 6$ , $BC = 3$ , point $P$ is chosen on $AB $ such that $\angle APD = 2 \angle CPB$. Find $AP$.

Published on 16 Apr 2026 - 10:01

score 288 · Answer 2 · 2026-04-11 20:22:15

288

Views

$ABCD$ is a rectangle with $AD = 2$ , $AB = 4$. $P$ is on $AB$ such that $AP : PB = 2 : 1$. $CE \perp DP$ at $E$. Find $CE$.

Published on 11 Apr 2026 - 20:22

score 599 · Answer 3 · 2026-03-25 19:10:33

599

Views

In the diagram, $AB$ $||$ $EF$ $||$ $DC$ . Given that $AC + BD = 250$ , $BC = 100$ and $EC + ED = 150$, find $CF$.

Published on 25 Mar 2026 - 19:10

score 780 · Answer 4 · 2026-04-16 11:56:49

780

Views

In the following figure, $AD = AB$. Also $\angle DAB = \angle DCB = \angle AEC = 90^\circ$ and $AE = 5$. Find the area of quadrilateral $ABCD$.

Published on 16 Apr 2026 - 11:56

score 573 · Answer 5 · 2026-04-14 07:38:52

573

Views

If $\angle p + \angle q + \angle r + \angle s + \angle t = 500^\circ$, find $\angle A + \angle B + \angle C + \angle D + \angle E$.

Published on 14 Apr 2026 - 7:38

score 896 · Answer 6 · 2026-04-13 19:28:09

896

Views

Let $ABCDEF$ be a hexagon such that the diagonals $AD,BE,CF$ intersect at point $O$.

Published on 13 Apr 2026 - 19:28

score 346 · Answer 7 · 2026-03-25 22:05:06

346

Views

Let $ABCD$ be a square. Points $A,B,G$ are collinear. $AC$ and $DG$ meet at $E$ , and $DG$ and $BC$ meet at $F$.

Published on 25 Mar 2026 - 22:05

score 447 · Answer 8 · 2026-04-13 17:21:12

447

Views

Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel.

Published on 13 Apr 2026 - 17:21

score 69 · Answer 9 · 2026-04-16 02:29:42

69

Views

$ABCD$ is a trapezium where $AB$ $||$ $CD$. Suppose $[\Delta BOC] = \frac{25}{9}$, and let $AB = b$, $CD = a$ where $a < b$.

Published on 16 Apr 2026 - 2:29

score 38 · Answer 10 · 2026-04-11 03:12:49

38

Views

Solving equation with natural numbers

Published on 11 Apr 2026 - 3:12

score 278 · Answer 11 · 2026-04-16 05:27:39

278

Views

$ABCD$ is a trapezium where $AB$ || $CD.$ Let $AB = b$ , $CD = a$ where $a < b$. Let $S$ be the area of trapezium $ABCD$.

Published on 16 Apr 2026 - 5:27

score 56 · Answer 12 · 2026-04-16 04:07:07

56

Views

How can one find the value of x such that $\left(1-\sqrt{1-(R-1)^2}\right)-\left(1-R^{\frac{1}{2-x}}\right)^{2-x}=0$, $(0<R<1)$?

Published on 16 Apr 2026 - 4:07

score 479 · Answer 13 · 2026-04-12 14:10:28

479

Views

Doubt in a 'probability problem

Published on 12 Apr 2026 - 14:10

score 108 · Answer 14 · 2026-04-15 06:31:49

108

Views

Finding $abc$ for distinct reals satisfying $\frac{a}{1+ab}=\frac{b}{1+bc}=\frac{c}{1+ca}$

Published on 15 Apr 2026 - 6:31

score 506 · Answer 15 · 2026-04-10 18:19:02

506

Views

Is my method of solving equation correct?

Published on 10 Apr 2026 - 18:19

List Question

In a rectangle $ABCD$ where $AB = 6$ , $BC = 3$ , point $P$ is chosen on $AB $ such that $\angle APD = 2 \angle CPB$. Find $AP$.

$ABCD$ is a rectangle with $AD = 2$ , $AB = 4$. $P$ is on $AB$ such that $AP : PB = 2 : 1$. $CE \perp DP$ at $E$. Find $CE$.

In the diagram, $AB$ $||$ $EF$ $||$ $DC$ . Given that $AC + BD = 250$ , $BC = 100$ and $EC + ED = 150$, find $CF$.

In the following figure, $AD = AB$. Also $\angle DAB = \angle DCB = \angle AEC = 90^\circ$ and $AE = 5$. Find the area of quadrilateral $ABCD$.

If $\angle p + \angle q + \angle r + \angle s + \angle t = 500^\circ$, find $\angle A + \angle B + \angle C + \angle D + \angle E$.

Let $ABCDEF$ be a hexagon such that the diagonals $AD,BE,CF$ intersect at point $O$.

Let $ABCD$ be a square. Points $A,B,G$ are collinear. $AC$ and $DG$ meet at $E$ , and $DG$ and $BC$ meet at $F$.

Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel.

$ABCD$ is a trapezium where $AB$ $||$ $CD$. Suppose $[\Delta BOC] = \frac{25}{9}$, and let $AB = b$, $CD = a$ where $a < b$.

Solving equation with natural numbers

$ABCD$ is a trapezium where $AB$ || $CD.$ Let $AB = b$ , $CD = a$ where $a < b$. Let $S$ be the area of trapezium $ABCD$.

How can one find the value of x such that $\left(1-\sqrt{1-(R-1)^2}\right)-\left(1-R^{\frac{1}{2-x}}\right)^{2-x}=0$, $(0<R<1)$?

Doubt in a 'probability problem

Finding $abc$ for distinct reals satisfying $\frac{a}{1+ab}=\frac{b}{1+bc}=\frac{c}{1+ca}$

Is my method of solving equation correct?

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