If the sides of the triangle are in the ratio $1: \sqrt{2}: 1,$ show that is a right $-$ angled triangle.
View full solution →Question types
Pythagoras Theorem [Proof and Simple Applications with Converse] question types
41 questions across 4 question groups — pick any mix to generate a MATHEMATICS paper with step-by-step answer keys.
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Sample Questions
Pythagoras Theorem [Proof and Simple Applications with Converse] questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
In equilateral $\triangle ABC, AD \perp BC$ and $BC = x \ cm.$ Find, in terms of $x,$ the length of $AD.$
View full solution →In the following figure, $AD$ is perpendicular to $BC$ and $D$ divides $BC$ in the ratio $1: 3$.
Prove that: $2 AC ^2=2 AB ^2+ BC ^2$
View full solution →Prove that: $2 AC ^2=2 AB ^2+ BC ^2$
In figure $AB=BC$ and $AD$ is perpendicular to $CD$.
Prove that: $AC ^2=2BC . DC$.
View full solution →Prove that: $AC ^2=2BC . DC$.
In a quadrilateral $\text{ABCD}, \angle B=90^{\circ}$ and $\angle D=90^{\circ}$.Prove that: $2 A C^2-A B^2=B C^2+C D^2+D A^2$
View full solution →In $\triangle ABC,\angle B = 90^o$ and $D$ is the mid$-$point of $BC.$Prove that: $AC^2= AD^2+ 3CD^2.$
View full solution →In the figure, given below, $A D \perp B C$.Prove that: $c^2=a^2+b^2-2 a x$.
View full solution →In $\triangle ABC, AB = AC$ and $BD$ is perpendicular to $AC.$Prove that: $BD^2- CD^2= 2CD \times AD.$
View full solution →Two poles of heights $6\ m$ and $11\ m$ stand vertically on a plane ground. If the distance between their feet is $12\ m;$find the distance between their tips.
View full solution →In the following figure, $O P, O Q$ and $O R$ are drawn perpendiculars to the sides $B C, C A$ and $A B$ respectively of $\triangle A B C$.Prove that: $AR ^2+ BP ^2+ CQ ^2= AQ ^2+ CP ^2+ BR ^2$
View full solution →$O$ is any point inside a rectangle $\text{ABCD}$.Prove that: $OB^2+ OD^2= OC^2+ OA^2$.
View full solution →In a rectangle $\text{ABCD}$,prove that: $AC^2+ BD^2= AB^2+ BC^2+ CD^2+ DA^2$.
View full solution →$M$ and $N$ are the mid$-$points of the sides $QR$ and $PQ$ respectively of a $\text{PQR}$, right$-$angled at $Q.$Prove that:$(i) PM^2+ RN^2= 5 MN^2,(ii) 4 PM^2= 4 PQ^2+ QR^2,$$(iii) 4 RN^2= PQ^2+ 4 QR^2(iv) 4 (PM^2+ RN^2) = 5 PR^2$
View full solution →$\text{ABC}$ is a triangle, right$-$angled at $B. M$ is a point on $BC$.Prove that: $AM^2+ BC^2= AC^2+ BM^2$.
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