Question

In a rhombus ABCD, the altitude from D to the
side AB bisects AB. Find the angles of rhombus.
​

Answer 1

From the figure consider DP bisect the side at point P

☯ Construct a line at BD

Consider ∆AMD and ∆BMD

WKT,

$\sf \: AM = BM ( M \: \: is \: \: the \: \: mid \: \: point \: \: of \: \: AB)$

$\sf \: \angle AMD= \angle \: BMD (90 \degree)$

$\sf \triangle \: AMD \cong \: \triangle \: BMD(By \: \: SAS \: \: Criteria)$

$\bf \red{The \: \: sides \: \: of \: \: a \: \: rhombus \: \: are \: \: equal}$

AB =AD

It can be written as

AD=AB=BD

In ∆ADB is an equilateral triangle

$\therefore \sf{ \angle \: A=60 \degree}$

$\bf \green{ \angle \: C= \angle \: A = 60 \degree}(in \: \: rhombus \: \: opposite \: \: angles \: \: are \: \: equal)$

we know that

$\tt \: \angle \: B+ \angle \: Q = 180 \degree$

$\tt \: \angle \: B = 180 \degree - 60 \degree$

$\tt \: \angle \: B = 120 \degree$

we know that,

$\bf \green{ \angle \: D= \angle \: B=120\degree}$

$\therefore \tt \pink{The \: \: angles \: \: of \: \: the \: \: rhombus \: \: is \angle \: C= \angle \: A = 60 \degree \: and \: \: \angle \: D= \angle \: B=120\degree}$

Answer 2

• From the figure consider DP bisect the side at point P

☯ Construct a line at BD

• Consider ∆AMD and ∆BMD

$\bf \red{The \: \: sides \: \: of \: \: a \: \: rhombus \: \: are \: \: equal}$

• AB =AD

• It can be written as

• AD=AB=BD

• In ∆ADB is an equilateral triangle

$\therefore \sf{ \angle \: A=60 \degree}$

$\bf \green{ \angle \: C= \angle \: A = 60 \degree}(in \: \: rhombus \: \: opposite \: \: angles \: \: are \: \: equal)∠C=∠A=60°$

• we know that

$\tt \: \angle \: B+ \angle \: Q = 180 \degree∠B+∠Q=180°$

$\tt \: \angle \: B = 180 \degree - 60 \degree∠B=180°−60°$

$\tt \: \angle \: B = 120 \degree∠B=120°$

• we know that,

$\bf \green{ \angle \: D= \angle \: B=120\degree}$

$\therefore \tt \pink{The \: \: angles \: \: of \: \: the \: \: rhombus \: \: is \angle \: C= \angle \: A = 60 \degree \: and \: \: \angle \: D= \angle \: B=120\degree}$

ItzJanhvi 10❤