Question : In $\mathrm{\Delta ABC, \angle BAC = 90^{\circ}}$ and $\mathrm{AD}$ is drawn perpendicular to $\mathrm{BC}$. If $\mathrm{BD} = 7\;\mathrm{cm}$ and $\mathrm{CD }= 28\;\mathrm{cm}$, what is the length of $\mathrm{AD}$?
Option 1: $3.5 \text{ cm}$
Option 2: $7 \text{ cm}$
Option 3: $10.5 \text{ cm}$
Option 4: $14 \text{ cm}$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $14 \text{ cm}$
Solution :
In the right-angled $\triangle \mathrm{ABC}$, $\mathrm{AD}$ is the altitude on the hypotenuse $\mathrm{BC}$. According to the property of triangles, the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and each other. $\therefore\mathrm{AD^2 = BD × DC}$ $⇒\mathrm{AD = \sqrt{BD × DC}}$ $⇒\mathrm{AD = \sqrt{BD×DC} = \sqrt{7 \text{ cm}× 28 \text{ cm}} = \sqrt{196 \text{ cm}^2} = 14 \text{ cm}}$ Hence, the correct answer is $14 \text{ cm}$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : In $\triangle$ABC, $\angle$A = 90°, AD$\perp$BC and AD = BD = 2 cm. The length of CD is:
Question : In $\triangle$ABC, $\angle$BAC = 90º and AD is perpendicular to BC. If AD = 6 cm and BD = 4 cm, then the length of BC is:
Question : In $\triangle \mathrm{ABC}$, $\angle \mathrm{ABC} = 90^{\circ}$, $\mathrm{BP}$ is drawn perpendicular to $\mathrm{AC}$. If $\angle \mathrm{BAP} = 50^{\circ},$ what is the value of $\angle \mathrm{PBC}?$
Question : ABC is a right angle triangle and $\angle ABC = 90^{\circ}$. BD is perpendicular on the side AC. What is the value of $(BD)^2$?
Question : $ABC$ is a triangle and $D$ is a point on the side $BC$. If $BC = 16\mathrm{~cm}$, $BD = 11 \mathrm{~cm}$ and $\angle ADC = \angle BAC$, then the length of $AC$ is equal to:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile