Let $PQRS$ be a rhombus, all sides of the rhombus have equal length, and its diagonal $PR$ and $SQ$ are intersecting each other at a point $O$. Diagonals in rhombus bisect each other at $90^{\circ}$.So, $PO=(\frac{PR}{2})$$=\frac{16}{2}$$=8\ cm$And, $SO=(\frac{SQ}{2})$$=\frac{30}{2}$$=15\ cm$Then, consider the triangle POS and apply the Pythagoras Theorem, $PS^2=PO^2+SO^2$$PS^2=8^2+15^2$$PS^2=64+225$$PS^2=289$$PS=\sqrt{289}$$PS=17\ cm$Hence, ... Read More

Given: Angles $Q$ and $R$ of a $∆PQR$ are $25^{\circ}$ and $65^{\circ}$.To do: To write the truth of the following:$(i).\ PQ^2+QR^2=RP^2$$(ii).\ PQ^2+RP^2=QR^2$$(iii).\ RP^2+QR^2=PQ^2$Solution:$∠PQR+∠QRP +∠RPQ = 180°$            [By angle sum property of a triangle]$25^{\circ}+65^{\circ}+\angle RPQ=180^{\circ}$$90^{\circ} +\angle RPQ =180^{\circ}$$\angle RPQ = 180^{\circ}-90^{\circ}$$\angle RPQ = 90^{\circ}$Thus $\Delta PQR$ is a ... Read More

Given: A rectangle whose length is $40\ cm$ and a diagonal is $41\ cm$.To do: To find the perimeter of the rectangle.Solution:$PR=41\ cm$, $PQ=40\ cm$Let breadth $(QR)$ be $x\ cm$Now, in right-angled triangle PQR$(PR)^2=(RQ)^2+(PQ)^2$$\Rightarrow (41)^2=x^2+(40)^2$$\Rightarrow 1681=x^2+1600$$\Rightarrow 1681-1600=x^2$$\Rightarrow \sqrt{81}=x$$\Rightarrow 9=x$Therefore, the breadth of the rectangle$=9\ cm$The perimeter of the rectange$=2(l+b)$$=2(40+9)$$=2(49)$$=98\ cm$. Read More

Given: $(i).\ 2.5 cm, \ 6.5 cm, \ 6 cm.$$(ii).\ 2 cm, \ 2 cm, \ 5 cm.$$(iii).\ 1.5 cm, \ 2cm, \ 2.5 cm.$To do: To write which of the above-given sides can be the sides of a right-angled triangle. In the case of right-angled triangles, the right angle has to ... Read More

Given: A tree is broken at a height of $5\ m$ from the ground and its top touches the ground at a distance of $12\ m$ from the base of the tree. To do: To find the original height of the tree.Solution:Let $ACB$ represent the tree before it breaks at the point $C$, ... Read More

Given:$PQR$ is a triangle, right-angled at P. $PQ  = 10\ cm $   ;    $PR = 24\ cm$To Find:  The value of $QR$.Solution:Since its aright angle triangle apply Pythagoras formula, angle P = 90°   ;  QR is hypotenuse.$QR^ 2    =  PQ^ 2    +   PR^2$$QR^ 2    =  10^2    +   24^2$$QR^ 2    =  ... Read More

Given :$ABC$ is a right-angled triangle at $C$.$AB= 25$ cm and $AC = 7$ cm.To do :We have to find the value of BC.Solution :Angle C is the right angle, this implies, that AB is the hypotenuse.Therefore, $AB^2=AC^2 + BC^2$$\Rightarrow (25)^2=(7)^2+BC^2$$\Rightarrow BC^2 = 625 - 49$$\Rightarrow BC^2 = 576$$\Rightarrow BC^2 = 24 \times ... Read More

Given: A $15\ m$ long ladder reached a window $12\ m$ high from the ground by placing it against a wall at a distance $a$. To do: To find the distance of the foot of the ladder from the wall. Solution:Let $AC$ be the ladder, and $A$ be the window $AC^2=AB^2+BC^2$ ... Read More

Given: $ABCD$ is quadrilateral. To do: To find whether $AB + BC + CD + DA Solution:The sum of the length of any two sides in a triangle should be greater than the length of the third side.In $\Delta AOB$,   $AB$+ob$ >In $\Delta BOC$,   $BC$+oc$ >In $\Delta COD$,   $CD$+od$ >In $\Delta AOD$,   ... Read More

Given: $ABCD$ is a quadrilateral.To do: To find whether $AB + BC + CD + DA > AC + BD?$Solution:The sum of the length of any two sides in a triangle should be greater than the length of the third side, ThereforeIn $\Delta ABC$,   $AB+BC>AC$    .......$(i)$$\Delta ADC$,      $AD+DC>AC$, ... Read More